Write the sample space and calculate the probability based on the given information.
(i) Two dice are rolled. What is the probability that the sum is a prime number greater than 5?
(ii) A bag contains 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?
(iii) Three coins. are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?
(iv) A four-digit number is formed using the digits 1, 2, 3, and 4 with no repetition. What is the probability that the number is even?
(v) A student takes a multiple-choice test with 3 questions, each having 4 options(A, B, C, D), with only one correct answer. What is the probability that the student guesses and gets exactly 2 answers correct?

Each die has outcomes: \(\{1,2,3,4,5,6\}\)
Total outcomes, \(n(S) = 6 × 6 = 36\)
Prime numbers greater than 5 possible from two dice are \(7\) and \(11\).
Outcomes with sum \(7\) : \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\)
Number of outcomes \(= 6\)

Outcomes with sum \(11\) : \((5,6),(6,5)\)
Number of outcomes \(= 2\)

So, total favourable outcomes \(= 6 + 2 = 8\)
\(P(prime\,sum\,greater\,than\,5) = \frac{8}{36} = \frac{2}{9}\)
The probability that the sum is a prime number greater than 5 is \(\frac{2}{9}\).

Bag contains: 4 red, 3 green, and 2 blue balls.
Total balls \(= 9\)
Since two balls are drawn without replacement, after the first ball is taken, only 8 balls remain.
Total outcomes, \(n(S) = 9 × 8 = 72\)

Possible colour pairs are:
(i) Red and Green
(ii) Red and Blue
(iii) Green and Blue
For (i): Red and Green outcomes
Case A: Choose 1 red from 4 reds and 1 green from 3 greens
\(4 × 3 = 12\)
Case B: Choose 1 green from 3 greens and 1 red from 4 reds
\(3 × 4 = 12\)
So, total Red-Green outcomes \(= 12 + 12 = 24\)

For (ii): Red and Blue outcomes
Case A: Choose 1 red from 4 reds and 1 blue from 2 blues
\(4 × 2 = 8\)
Case B: Choose 1 blue from 2 blues and 1 red from 4 reds
\(2 × 4 = 8\)
So, total Red-Blue outcomes \(= 8 + 8 = 16\)

For (iii): Green and Blue outcomes
Case A: Choose 1 green from 3 greens and 1 blue from 2 blues
\(3 × 2 = 6\)
Case B: Choose 1 blue from 2 blues and 1 green from 3 greens
\(2 × 3 = 6\)
So, total Green-Blue outcomes \(= 6 + 6 = 12\)

Total favourable outcomes, \( = 24 + 16 + 12 = 52\)
Therefore, \(P(both\,balls\,are\,of\,different\,colours) = \frac{52}{72} = \frac{13}{18}\)
The probability that both balls are of different colours is \(\frac{13}{18}\).

Sample space \(= \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}\)
Total outcomes, \(n(S) = 8\)
Event, E: first coin shows heads and exactly two heads occur in total
Favourable outcomes: \(\{HHT, HTH\}\)
Total favourable outcomes \(= 2\)
Therefore, \(P(E) = \frac{2}{8} = \frac{1}{4}\)
The probability that the first coin shows heads and exactly two heads occur in total is \(\frac{1}{4}\).

Using all digits without repetition, we get
Total outcomes ,\(n(S) = 4! = 24\)

A number is even if the last digit is even. The possible last even digits here are 2 and 4.
Case A: Last digit 2
The remaining digits can be arranged in \(3! = 6\) ways.
Case B: Last digit 4
The remaining digits can be arranged in \(3! = 6\) ways.
Total favourable outcomes \(= 6 + 6 = 12\)
Therefore, \(P(number\,is\,even) = \frac{12}{24} = \frac{1}{2}\)
The probability that the number is even is \(\frac{1}{2}\).

There are three questions. Each question has: (i) 1 correct answer(C)
(ii) 3 wrong answers(W)
\(P(C) = \frac{1}{4}\)
\(P(W) = \frac{3}{4}\)

Event: Exactly 2 correct answers
Possible patterns : \(CCW, CWC, WCC\)
Total favourable outcomes \(= 3\)
Probability of one pattern(say \(CCW\)) : \(\frac{1}{4} × \frac{1}{4} × \frac{3}{4} = \frac{3}{64}\)
Therefore, Total Probability, \(P(E) = 3 × \frac{3}{64} = \frac{9}{64}\)
The probability that the student guesses and gets exactly 2 answers correct is \(\frac{9}{64}\).